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Two-Dimensional Elastic Collision of Equal Masses

  1. Mar 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that if an elastic collision between a mass and a stationary target of equal mass is not head-on that the projectile and target final velocities are perpendicular. (Hint: Square the conservation of momentum equation, using ##p^2=p\cdot p##, and compare the resulting equation with the energy conservation equation.)

    2. Relevant equations
    ##p=mv##
    ##K=\frac{1}{2}mv^2##

    3. The attempt at a solution
    So I followed the hint and got $$\overrightarrow{p_1^2}=\overrightarrow{p_1^{'2}}+\overrightarrow{p_2^{'2}}+2\left(\overrightarrow{p_1}\cdot \overrightarrow{p_2}\right)$$ Plugging in values I got $$m^2\overrightarrow{v_1^2}=m^2\overrightarrow{v_1^{'2}}+m^2\overrightarrow{v_2^{'2}}+2\left(m\overrightarrow{v}_1\cdot m\overrightarrow{v}_2\right)$$ $$\overrightarrow{v_1^2}=\overrightarrow{v_1^{'2}}+\overrightarrow{v_2^{'2}}$$ For the x- and y-components I got the following assuming the target traveled along the x-axis after the collision $$v_1^2cos^2(\theta)=v_1^{'2}cos^2(\theta ^{'})+v_2^{'2}$$ $$v_1^2sin^2(\theta)=v_1^{'2}sin^2(\theta ^{'})$$ This is where I got stuck and wasn't sure how to solve for ##v_1^{'}## or ##v_2^{'}##. I'm not entirely sure when or how to apply the equation for kinetic energy. Am I in the right direction or did I do something wrong?
     
    Last edited: Mar 26, 2017
  2. jcsd
  3. Mar 26, 2017 #2

    kuruman

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    You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.
     
  4. Mar 26, 2017 #3
    I forgot to square the initial velocity. Were there any other mistakes?
     
  5. Mar 26, 2017 #4
    I think I got it.

    So the momentum in the x and y-direction is $$v_1=v_1^{'}cos(\theta _1)+v_2^{'}cos(\theta _2)$$ $$0=v_1^{'}sin(\theta _1)-v_2^{'}sin(\theta _2)$$ I then used the following to find the magnitude of the momentum $$p^2=p_x^2+p_y^2$$ Doing this I got $$v_1=v_1^{'2}(cos^2(\theta _1)+sin^2(\theta _1))+v_2^{'2}(cos^2(\theta _2)+sin^2(\theta _2))+2v_1^{'}v_2^{'}(cos(\theta _1)cos(\theta _2)-sin(\theta _1)sin(\theta _2))$$ Using trig identities and the law of cosines, I simplified it to $$v_1^2=v_1^{'2}+v_2^{'2}+2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Knowing that kinetic energy is conserved in an elastic collision I used the equation for kinetic energy and simplified $$\frac{1}{2}mv_1^2=\frac{1}{2}mv_1^{'2}+\frac{1}{2}mv_2^{'2}$$ $$v_1^2=v_1^{'2}+v_2^{'2}$$ With this formula, ##v_1^2## can be substituted for ##v_1^{'2}+v_2^{'2}## After substituting the equation simplifies to $$0=2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$ Solving for the last equation results in $$\theta _1+\theta _2 = 90°\text{ or }\frac{\pi}{2}$$ Thus, the two velocities must be perpendicular.
     
  6. Mar 26, 2017 #5

    haruspex

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    it results in at least one of those three. How do you rule out the first two?
     
  7. Mar 26, 2017 #6
    The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.
     
  8. Mar 26, 2017 #7

    haruspex

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    You are given that the collision is not head-on. From that you can show neither velocity is zero.
     
  9. Mar 26, 2017 #8
    That makes sense, thank you very much.
     
  10. Mar 27, 2017 #9

    kuruman

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    Here is a quick geometric proof, not that your algebraic proof is incorrect.
    Momentum conservation for a two equal-mass collision requires that ##\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}##.
    Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
    Energy conservation for a two equal-mass collision requires that ##v_1^2= {v'_1}^2+{v'_1}^2##.
    Therefore, said triangle obeys the Pythagorean theorem with ##v_1## as the hypotenuse and ##{v'_1}## and ##{v'_2}## as the two right sides.
     
  11. Mar 27, 2017 #10
    Thank you! I wish I had seen that sooner.
     
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